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April 28th, 2015, 01:25 PM   #1
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Please help me to understand

I have recently stumbled across this following question:

The answers are the following:

I would like to know how to work out the answers as I got them wrong.
i) I understand why it would be xy, but I don't understand why if it is xy the number would be 2^p+s and not 4^p+s.
ii) I completely don't understand this.
iii) Completely confused with this one as well.

It would be great if someone could help and say how those answers came about.
Thanks a lot.

Last edited by skipjack; April 28th, 2015 at 03:57 PM.
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April 28th, 2015, 01:33 PM   #2
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i) $\displaystyle 2^{p+s}=(2^p)(2^s)=xy$

ii) $\displaystyle 2^{2p}=(2^p)^2=x^2$

iii) $\displaystyle 2^{p-1}=(2^p)(2^{-1})=2^p\cdot\frac12=x\cdot\frac12=\frac12x$
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April 28th, 2015, 01:50 PM   #3
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Recall that (for any positive integers $p$ and $s$)
$$2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} \\
2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$s$ terms}} $$
So when we multiply we get
$$2^p \times 2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$s$ terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p+s$ terms}} = 2^{p+s}$$

Similarly,
$$\left(2^p\right)^2 = 2^p \times 2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p+p$ terms}} = 2^{p+p} = 2^{2p}$$

Finally,
$$2^{p-1} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p-1$ terms}} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} = 2^p$$ so we can divide by 2 to get
$$2^{p-1} = \frac12 2^p$$

These results generalise to any $p$ and $s$.
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April 28th, 2015, 02:03 PM   #4
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Quote:
Originally Posted by greg1313 View Post
i) $\displaystyle 2^{p+s}=(2^p)(2^s)=xy$

ii) $\displaystyle 2^{2p}=(2^p)^2=x^2$

iii) $\displaystyle 2^{p-1}=(2^p)(2^{-1})=2^p\cdot\frac12=x\cdot\frac12=\frac12x$
Cheers for the working out, very helpful

Quote:
Originally Posted by v8archie View Post
Recall that (for any positive integers $p$ and $s$)
$$2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} \\
2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$s$ terms}} $$
So when we multiply we get
$$2^p \times 2^s = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$s$ terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p+s$ terms}} = 2^{p+s}$$

Similarly,
$$\left(2^p\right)^2 = 2^p \times 2^p = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} \times \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p+p$ terms}} = 2^{p+p} = 2^{2p}$$

Finally,
$$2^{p-1} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p-1$ terms}} \times 2 = \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{\text{$p$ terms}} = 2^p$$ so we can divide by 2 to get
$$2^{p-1} = \frac12 2^p$$

These results generalise to any $p$ and $s$.
I understand now, very well explained
thank you very much (:
One more thing, as you said you explained the p+s terms come together to form 2^p+s
But how come when you do a multiplication such as 9^3 x 9^3 = It equals 81^6 and not 9^6

Last edited by nautudent; April 28th, 2015 at 02:07 PM.
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April 28th, 2015, 02:26 PM   #5
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Quote:
Originally Posted by nautudent View Post
But how come when you do a multiplication such as 9^3 x 9^3 = It equals 81^6 and not 9^6
$$9^3 \times 9^3= (9 \times 9 \times 9) \times ( 9 \times 9 \times 9) = \underbrace{9 \times 9 \times 9 \times 9 \times 9 \times 9}_{\text{$6$ terms}} = 9^6$$
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May 2nd, 2015, 05:08 AM   #6
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Quote:
Originally Posted by v8archie View Post
$$9^3 \times 9^3= (9 \times 9 \times 9) \times ( 9 \times 9 \times 9) = \underbrace{9 \times 9 \times 9 \times 9 \times 9 \times 9}_{\text{$6$ terms}} = 9^6$$
Hmm so it is 9^6
so why would this be right??

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May 2nd, 2015, 05:34 AM   #7
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$$\begin{aligned}4pq \times 6p^2q &= 4 \times p \times q \times 6 \times p \times p \times q \\ &= \underbrace{4 \times 6}_{24} \times \underbrace{p \times p \times p}_{p^3} \times \underbrace{q \times q}_{q^2} \\ &= 24p^3q^2\end{aligned}$$
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May 2nd, 2015, 05:44 AM   #8
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One can add the exponents when finding the product of powers of the same thing: $p^mp^n = p^{m+n}$.

If the exponents are equal, one can find the product as follows: $p^nq^n = (pq)^n$.

Hence $9^39^3$ is equal to both $9^6$ and $81^3$.
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